Fermi Questions Marathon

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OldSpice
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Re: Fermi Questions Marathon

Postby OldSpice » January 22nd, 2012, 12:06 pm

I'd say I hit about -E1 lights a day with -E2 days a year and -E1 year lifetime meaning I wait at -E4 lights during my life. There are about -E5 total minutes in a lifetime and if every redlight lasts approximately 2 minutes, the average person would spend -E4/-E5 (1/10) of their life waiting at stopligts.

How many metal track cleats are used in the U.S. each year.
That's a tricky one.
The average track team has E1 members, There are E5 high schools and E4 colleges.
I'm probably really off but I'd say around E7.

If you take every book sold in the world in the past ten years and stack them vertically, how high would the stack be in terameters?
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Re: Fermi Questions Marathon

Postby hmcginny » January 22nd, 2012, 1:24 pm

well there are 7E9 people in the world, and each year each person probably buys E1 books, but we'll round it up to a low E2 because of schools and libraries. so thats E12 books per year, and E13 for 10 years. Each book is probably around an inch thick, so E13 inches tall, which is 2E13 centimeters or 2E-1 terameters so E-1 is my answer.

How many pages of paper will be used to print the midterm exams for every class in every US high school?
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Re: Fermi Questions Marathon

Postby A123456789 » January 22nd, 2012, 2:59 pm

There are about 20 million high school students in the U.S. Suppose that each student takes 4 exams on average and that the average midterm uses 20 pages of paper. 2E7X4X20=160E7=9.

How many grains of sand would it take to fill an Olympic swimming pool?

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Re: Fermi Questions Marathon

Postby ndclscienceolympiad » January 25th, 2012, 10:32 pm

So I'll say a grain of sand has a rough volume of 2E-2 cm^3 which converts to 2E-8 m^3. An olympic swimming pool is 2,500m^3 in volume, so 2.5E3. 2.5E3/2E-8 = 1.25E11------>[11]


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Re: Fermi Questions Marathon

Postby Schrodingerscat » January 26th, 2012, 6:39 am

I will simply guess 10^3, as I am guessing around 1000lbs of beef, although it could round up to 10^4.

How many atoms of helium are in the average classroom?

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Re: Fermi Questions Marathon

Postby danny9696 » January 31st, 2012, 8:59 am

"How many atoms of Helium are in the average classroom?"

A classroom is about 3x10x10 = 300 meters cubed = 3E8 cm^3
Liter = cm^3 = 3E8 liters
PV = nRT, assume P and R to be 1: 3E8 = 3E2K (n)
n=1E6 moles of all gases
Helium is BARELY present in air, probably something parts per million so 1E6/1E6 = 1 mol of helium
6E23 atoms, therefore around 24 <- fermi answer

Question: How long would it take for light from Alpha Centauri to reach the Earth in minutes?

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Re: Fermi Questions Marathon

Postby hmcginny » January 31st, 2012, 9:23 am

Speed of light is 3E8 m/s, distance to alpha tauri is probably E12 meters (just a guess, but its probably within 1 or 2), so it would take .33E4 or 3E3 seconds, which is 5E1 minutes, which is on the border, but I'll assume alpha tauri is on the lower end of E12 meters so lets go with E1 minutes.

How many hairs are on a typical house cat?
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Re: Fermi Questions Marathon

Postby AlphaTauri » January 31st, 2012, 12:42 pm

Eh, I'd have to disagree there, hmcginny. Alpha Centauri is about 4.3 light years from Earth, so it'd take the light 4.3 years to reach us - which converts to 4.3 years * 365.25 days * 24 hours * 60 minutes = 2.3E6 minutes.

Or the short way - E0 years * E2 days/year * E1 hours/day * E2 minutes/hour = E5, which is pretty close (not sure where/if I should round up to make it E6, as I don't do Fermi).
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Re: Fermi Questions Marathon

Postby hmcginny » January 31st, 2012, 3:34 pm

well looks like my estimate was way off, if i had started with the right number (E17 meters) I would have gotten it. whoops. question still stands about how many hairs there are on a typical house cat though.
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Re: Fermi Questions Marathon

Postby aditya8081 » February 7th, 2012, 11:01 am

Lets estimate that a standard housecat is a cylinder of radius 10 cm and a length of 30 cm.
This means that the Surface area of a house cat is about 2*pi*10 cm* 30cm + 2*pi* (10 cm)^2, which comes out to 60 cm^2 + 600 cm^2, so Surface Area= 7e2 cm^2 The average follicle density of a cat is about 10 hairs / sq cm, so 7e3 hairs, meaning my answer will be 4

Now my Fermi Question:
If all of cornfields in the US were used to make popcorn for one year, how massive would the resulting popcorn be?


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